JEE Challenger
More from p-Block Elements (Group 15-18)
ChemMedAdv 2009 • Paper 1

Identification of Compound X in Methylene Blue Synthesis

Comprehension Passage

pp-Amino-N,NN,N-dimethylaniline is added to a strongly acidic solution of X\mathbf{X}. The resulting solution is treated with a few drops of aqueous solution of Y\mathbf{Y} to yield blue coloration due to the formation of methylene blue. Treatment of the aqueous solution of Y\mathbf{Y} with the reagent potassium hexacyanoferrate(II) leads to the formation of an intense blue precipitate. The precipitate dissolves on excess addition of the reagent. Similarly, treatment of the solution of Y\mathbf{Y} with the solution of potassium hexacyanoferrate(III) leads to a brown coloration due to the formation of Z\mathbf{Z}.

The compound X\mathbf{X} is

Options

A

NaNO3\text{NaNO}_3

B

NaCl\text{NaCl}

C

Na2SO4\text{Na}_2\text{SO}_4

D

Na2S\text{Na}_2\text{S}

Correct

Topics & Concepts

p-Block Elements (Group 15-18)p-Block Elements

Step-by-Step Solution

To determine the identity of compound X\mathbf{X}, we analyze the reactions described in the passage:

  1. Identification of Y\mathbf{Y} and Z\mathbf{Z}:

    • Treatment of an aqueous solution of Y\mathbf{Y} with potassium hexacyanoferrate(II), K4[Fe(CN)6]\text{K}_4[\text{Fe}(\text{CN})_6], gives an intense blue precipitate (Prussian blue):
      4Fe3++3[Fe(CN)6]4−⟶Fe4[Fe(CN)6]3↓ (Prussian blue)4\text{Fe}^{3+} + 3[\text{Fe}(\text{CN})_6]^{4-} \longrightarrow \text{Fe}_4[\text{Fe}(\text{CN})_6]_3 \downarrow \text{ (Prussian blue)}
    • Treatment of Y\mathbf{Y} with potassium hexacyanoferrate(III), K3[Fe(CN)6]\text{K}_3[\text{Fe}(\text{CN})_6], yields a brown coloration due to the formation of Fe[Fe(CN)6]\text{Fe}[\text{Fe}(\text{CN})_6] (Z\mathbf{Z}):
      Fe3++[Fe(CN)6]3−⟶Fe[Fe(CN)6] (Brown)\text{Fe}^{3+} + [\text{Fe}(\text{CN})_6]^{3-} \longrightarrow \text{Fe}[\text{Fe}(\text{CN})_6] \text{ (Brown)}
    • Therefore, the compound Y\mathbf{Y} contains Fe3+\text{Fe}^{3+} ions (specifically, Y=FeCl3\mathbf{Y} = \text{FeCl}_3).

  2. Identification of X\mathbf{X} (Methylene Blue Test):

    • The formation of methylene blue occurs when pp-amino-N,NN,N-dimethylaniline (N,NN,N-dimethyl-pp-phenylenediamine) reacts with a sulfide (S2−\text{S}^{2-}) source in a strongly acidic medium in the presence of an oxidizing agent (Fe3+\text{Fe}^{3+}):
      2 (p-NH2-C6H4-N(CH3)2)+S2−+6Fe3+⟶Methylene Blue+6Fe2++NH4++4H+2\,(p\text{-NH}_2\text{-C}_6\text{H}_4\text{-N}(\text{CH}_3)_2) + \text{S}^{2-} + 6\text{Fe}^{3+} \longrightarrow \text{Methylene Blue} + 6\text{Fe}^{2+} + \text{NH}_4^+ + 4\text{H}^+
    • This reaction serves as a standard qualitative and colorimetric test for the detection of sulfide ions (S2−\text{S}^{2-}).
    • Among the given options, the source of sulfide ions is sodium sulfide, Na2S\text{Na}_2\text{S}.

Therefore, the compound X\mathbf{X} is Na2S\text{Na}_2\text{S}.

Correct Option: D

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